3.1475 \(\int \frac{\csc (c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=402 \[ \frac{2 b^3 \left (3 a^2-b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a d \left (a^2-b^2\right )^{7/2}}+\frac{b^3 \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a d \left (a^2-b^2\right )^{7/2}}+\frac{2 b^3 \left (-3 a^2 b^2+6 a^4+b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 d \left (a^2-b^2\right )^{7/2}}+\frac{b^4 \left (3 a^2-b^2\right ) \cos (c+d x)}{a^2 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac{3 b^4 \cos (c+d x)}{2 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac{b^4 \cos (c+d x)}{2 a d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{\cos (c+d x)}{2 d (a+b)^3 (1-\sin (c+d x))}+\frac{\cos (c+d x)}{2 d (a-b)^3 (\sin (c+d x)+1)} \]

[Out]

(2*b^3*(3*a^2 - b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a*(a^2 - b^2)^(7/2)*d) + (b^3*(2*a^2 +
b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a*(a^2 - b^2)^(7/2)*d) + (2*b^3*(6*a^4 - 3*a^2*b^2 +
b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*(a^2 - b^2)^(7/2)*d) - ArcTanh[Cos[c + d*x]]/(a^3*
d) + Cos[c + d*x]/(2*(a + b)^3*d*(1 - Sin[c + d*x])) + Cos[c + d*x]/(2*(a - b)^3*d*(1 + Sin[c + d*x])) + (b^4*
Cos[c + d*x])/(2*a*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x])^2) + (3*b^4*Cos[c + d*x])/(2*(a^2 - b^2)^3*d*(a + b*Si
n[c + d*x])) + (b^4*(3*a^2 - b^2)*Cos[c + d*x])/(a^2*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.509691, antiderivative size = 402, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2897, 3770, 2648, 2664, 2754, 12, 2660, 618, 204} \[ \frac{2 b^3 \left (3 a^2-b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a d \left (a^2-b^2\right )^{7/2}}+\frac{b^3 \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a d \left (a^2-b^2\right )^{7/2}}+\frac{2 b^3 \left (-3 a^2 b^2+6 a^4+b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 d \left (a^2-b^2\right )^{7/2}}+\frac{b^4 \left (3 a^2-b^2\right ) \cos (c+d x)}{a^2 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac{3 b^4 \cos (c+d x)}{2 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac{b^4 \cos (c+d x)}{2 a d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{\cos (c+d x)}{2 d (a+b)^3 (1-\sin (c+d x))}+\frac{\cos (c+d x)}{2 d (a-b)^3 (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]*Sec[c + d*x]^2)/(a + b*Sin[c + d*x])^3,x]

[Out]

(2*b^3*(3*a^2 - b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a*(a^2 - b^2)^(7/2)*d) + (b^3*(2*a^2 +
b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a*(a^2 - b^2)^(7/2)*d) + (2*b^3*(6*a^4 - 3*a^2*b^2 +
b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*(a^2 - b^2)^(7/2)*d) - ArcTanh[Cos[c + d*x]]/(a^3*
d) + Cos[c + d*x]/(2*(a + b)^3*d*(1 - Sin[c + d*x])) + Cos[c + d*x]/(2*(a - b)^3*d*(1 + Sin[c + d*x])) + (b^4*
Cos[c + d*x])/(2*a*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x])^2) + (3*b^4*Cos[c + d*x])/(2*(a^2 - b^2)^3*d*(a + b*Si
n[c + d*x])) + (b^4*(3*a^2 - b^2)*Cos[c + d*x])/(a^2*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x]))

Rule 2897

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x]
/; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && (LtQ[m, -1] || (EqQ[m, -1] && G
tQ[p, 0]))

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\csc (c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\int \left (\frac{\csc (c+d x)}{a^3}-\frac{1}{2 (a+b)^3 (-1+\sin (c+d x))}-\frac{1}{2 (a-b)^3 (1+\sin (c+d x))}-\frac{b^3}{a \left (-a^2+b^2\right ) (a+b \sin (c+d x))^3}+\frac{b^3 \left (3 a^2-b^2\right )}{a^2 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac{6 a^4 b^3-3 a^2 b^5+b^7}{a^3 \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}\right ) \, dx\\ &=\frac{\int \csc (c+d x) \, dx}{a^3}-\frac{\int \frac{1}{1+\sin (c+d x)} \, dx}{2 (a-b)^3}-\frac{\int \frac{1}{-1+\sin (c+d x)} \, dx}{2 (a+b)^3}+\frac{b^3 \int \frac{1}{(a+b \sin (c+d x))^3} \, dx}{a \left (a^2-b^2\right )}+\frac{\left (b^3 \left (3 a^2-b^2\right )\right ) \int \frac{1}{(a+b \sin (c+d x))^2} \, dx}{a^2 \left (a^2-b^2\right )^2}+\frac{\left (b^3 \left (6 a^4-3 a^2 b^2+b^4\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{a^3 \left (a^2-b^2\right )^3}\\ &=-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}+\frac{\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}+\frac{b^4 \cos (c+d x)}{2 a \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac{b^4 \left (3 a^2-b^2\right ) \cos (c+d x)}{a^2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac{b^3 \int \frac{-2 a+b \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )^2}+\frac{\left (b^3 \left (3 a^2-b^2\right )\right ) \int \frac{a}{a+b \sin (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )^3}+\frac{\left (2 b^3 \left (6 a^4-3 a^2 b^2+b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 \left (a^2-b^2\right )^3 d}\\ &=-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}+\frac{\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}+\frac{b^4 \cos (c+d x)}{2 a \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac{3 b^4 \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac{b^4 \left (3 a^2-b^2\right ) \cos (c+d x)}{a^2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac{b^3 \int \frac{2 a^2+b^2}{a+b \sin (c+d x)} \, dx}{2 a \left (a^2-b^2\right )^3}+\frac{\left (b^3 \left (3 a^2-b^2\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{a \left (a^2-b^2\right )^3}-\frac{\left (4 b^3 \left (6 a^4-3 a^2 b^2+b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 \left (a^2-b^2\right )^3 d}\\ &=\frac{2 b^3 \left (6 a^4-3 a^2 b^2+b^4\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{7/2} d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}+\frac{\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}+\frac{b^4 \cos (c+d x)}{2 a \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac{3 b^4 \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac{b^4 \left (3 a^2-b^2\right ) \cos (c+d x)}{a^2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac{\left (b^3 \left (2 a^2+b^2\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{2 a \left (a^2-b^2\right )^3}+\frac{\left (2 b^3 \left (3 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a \left (a^2-b^2\right )^3 d}\\ &=\frac{2 b^3 \left (6 a^4-3 a^2 b^2+b^4\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{7/2} d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}+\frac{\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}+\frac{b^4 \cos (c+d x)}{2 a \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac{3 b^4 \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac{b^4 \left (3 a^2-b^2\right ) \cos (c+d x)}{a^2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac{\left (4 b^3 \left (3 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a \left (a^2-b^2\right )^3 d}+\frac{\left (b^3 \left (2 a^2+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a \left (a^2-b^2\right )^3 d}\\ &=\frac{2 b^3 \left (3 a^2-b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{7/2} d}+\frac{2 b^3 \left (6 a^4-3 a^2 b^2+b^4\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{7/2} d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}+\frac{\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}+\frac{b^4 \cos (c+d x)}{2 a \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac{3 b^4 \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac{b^4 \left (3 a^2-b^2\right ) \cos (c+d x)}{a^2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac{\left (2 b^3 \left (2 a^2+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a \left (a^2-b^2\right )^3 d}\\ &=\frac{2 b^3 \left (3 a^2-b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{7/2} d}+\frac{b^3 \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{7/2} d}+\frac{2 b^3 \left (6 a^4-3 a^2 b^2+b^4\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{7/2} d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}+\frac{\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}+\frac{b^4 \cos (c+d x)}{2 a \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac{3 b^4 \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac{b^4 \left (3 a^2-b^2\right ) \cos (c+d x)}{a^2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 6.58863, size = 322, normalized size = 0.8 \[ \frac{9 a^2 b^4 \cos (c+d x)-2 b^6 \cos (c+d x)}{2 a^2 d (a-b)^3 (a+b)^3 (a+b \sin (c+d x))}+\frac{b^3 \left (-7 a^2 b^2+20 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (a \sin \left (\frac{1}{2} (c+d x)\right )+b \cos \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2-b^2}}\right )}{a^3 d \left (a^2-b^2\right )^{7/2}}+\frac{\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 d}-\frac{\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 d}+\frac{b^4 \cos (c+d x)}{2 a d (a-b)^2 (a+b)^2 (a+b \sin (c+d x))^2}+\frac{\sin \left (\frac{1}{2} (c+d x)\right )}{d (a+b)^3 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{\sin \left (\frac{1}{2} (c+d x)\right )}{d (a-b)^3 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[c + d*x]*Sec[c + d*x]^2)/(a + b*Sin[c + d*x])^3,x]

[Out]

(b^3*(20*a^4 - 7*a^2*b^2 + 2*b^4)*ArcTan[(Sec[(c + d*x)/2]*(b*Cos[(c + d*x)/2] + a*Sin[(c + d*x)/2]))/Sqrt[a^2
- b^2]])/(a^3*(a^2 - b^2)^(7/2)*d) - Log[Cos[(c + d*x)/2]]/(a^3*d) + Log[Sin[(c + d*x)/2]]/(a^3*d) + Sin[(c +
d*x)/2]/((a + b)^3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - Sin[(c + d*x)/2]/((a - b)^3*d*(Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2])) + (b^4*Cos[c + d*x])/(2*a*(a - b)^2*(a + b)^2*d*(a + b*Sin[c + d*x])^2) + (9*a^2*b^4*Cos
[c + d*x] - 2*b^6*Cos[c + d*x])/(2*a^2*(a - b)^3*(a + b)^3*d*(a + b*Sin[c + d*x]))

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Maple [B]  time = 0.178, size = 787, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)*sec(d*x+c)^2/(a+b*sin(d*x+c))^3,x)

[Out]

-1/d/(a+b)^3/(tan(1/2*d*x+1/2*c)-1)+11/d*b^5/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)
^2*tan(1/2*d*x+1/2*c)^3-4/d*b^7/(a-b)^3/(a+b)^3/a^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/
2*d*x+1/2*c)^3+10/d*a/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^2
*b^4+17/d*b^6/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/a*tan(1/2*d*x+1/2*c)^2-6/d*b
^8/(a-b)^3/(a+b)^3/a^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^2+29/d*b^5/(a-b)
^3/(a+b)^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)-8/d*b^7/(a-b)^3/(a+b)^3/a^2/
(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)+10/d*b^4/(a-b)^3/(a+b)^3/(tan(1/2*d*x+1
/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*a-3/d*b^6/(a-b)^3/(a+b)^3/a/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)
*b+a)^2+20/d*a/(a-b)^3/(a+b)^3*b^3/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))-7/
d*b^5/(a-b)^3/(a+b)^3/a/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))+2/d*b^7/(a-b)
^3/(a+b)^3/a^3/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))+1/d/(a-b)^3/(tan(1/2*d
*x+1/2*c)+1)+1/d/a^3*ln(tan(1/2*d*x+1/2*c))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*sec(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 19.155, size = 3615, normalized size = 8.99 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*sec(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

[-1/4*(4*a^10 - 12*a^8*b^2 + 12*a^6*b^4 - 4*a^4*b^6 + 2*(10*a^8*b^2 - 2*a^6*b^4 - 11*a^4*b^6 + 3*a^2*b^8)*cos(
d*x + c)^2 - ((20*a^4*b^5 - 7*a^2*b^7 + 2*b^9)*cos(d*x + c)^3 - 2*(20*a^5*b^4 - 7*a^3*b^6 + 2*a*b^8)*cos(d*x +
c)*sin(d*x + c) - (20*a^6*b^3 + 13*a^4*b^5 - 5*a^2*b^7 + 2*b^9)*cos(d*x + c))*sqrt(-a^2 + b^2)*log(-((2*a^2 -
b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(
-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) + 2*((a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*
a^2*b^8 + b^10)*cos(d*x + c)^3 - 2*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*cos(d*x + c)*sin(d*x +
c) - (a^10 - 3*a^8*b^2 + 2*a^6*b^4 + 2*a^4*b^6 - 3*a^2*b^8 + b^10)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) -
2*((a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b^10)*cos(d*x + c)^3 - 2*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4
*a^3*b^7 + a*b^9)*cos(d*x + c)*sin(d*x + c) - (a^10 - 3*a^8*b^2 + 2*a^6*b^4 + 2*a^4*b^6 - 3*a^2*b^8 + b^10)*co
s(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(2*a^9*b - 6*a^7*b^3 + 6*a^5*b^5 - 2*a^3*b^7 - (6*a^7*b^3 + 5*a^5
*b^5 - 13*a^3*b^7 + 2*a*b^9)*cos(d*x + c)^2)*sin(d*x + c))/((a^11*b^2 - 4*a^9*b^4 + 6*a^7*b^6 - 4*a^5*b^8 + a^
3*b^10)*d*cos(d*x + c)^3 - 2*(a^12*b - 4*a^10*b^3 + 6*a^8*b^5 - 4*a^6*b^7 + a^4*b^9)*d*cos(d*x + c)*sin(d*x +
c) - (a^13 - 3*a^11*b^2 + 2*a^9*b^4 + 2*a^7*b^6 - 3*a^5*b^8 + a^3*b^10)*d*cos(d*x + c)), -1/2*(2*a^10 - 6*a^8*
b^2 + 6*a^6*b^4 - 2*a^4*b^6 + (10*a^8*b^2 - 2*a^6*b^4 - 11*a^4*b^6 + 3*a^2*b^8)*cos(d*x + c)^2 + ((20*a^4*b^5
- 7*a^2*b^7 + 2*b^9)*cos(d*x + c)^3 - 2*(20*a^5*b^4 - 7*a^3*b^6 + 2*a*b^8)*cos(d*x + c)*sin(d*x + c) - (20*a^6
*b^3 + 13*a^4*b^5 - 5*a^2*b^7 + 2*b^9)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 -
b^2)*cos(d*x + c))) + ((a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b^10)*cos(d*x + c)^3 - 2*(a^9*b - 4*a^7*
b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*cos(d*x + c)*sin(d*x + c) - (a^10 - 3*a^8*b^2 + 2*a^6*b^4 + 2*a^4*b^6 - 3
*a^2*b^8 + b^10)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - ((a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b
^10)*cos(d*x + c)^3 - 2*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*cos(d*x + c)*sin(d*x + c) - (a^10
- 3*a^8*b^2 + 2*a^6*b^4 + 2*a^4*b^6 - 3*a^2*b^8 + b^10)*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - (2*a^9*b
- 6*a^7*b^3 + 6*a^5*b^5 - 2*a^3*b^7 - (6*a^7*b^3 + 5*a^5*b^5 - 13*a^3*b^7 + 2*a*b^9)*cos(d*x + c)^2)*sin(d*x +
c))/((a^11*b^2 - 4*a^9*b^4 + 6*a^7*b^6 - 4*a^5*b^8 + a^3*b^10)*d*cos(d*x + c)^3 - 2*(a^12*b - 4*a^10*b^3 + 6*
a^8*b^5 - 4*a^6*b^7 + a^4*b^9)*d*cos(d*x + c)*sin(d*x + c) - (a^13 - 3*a^11*b^2 + 2*a^9*b^4 + 2*a^7*b^6 - 3*a^
5*b^8 + a^3*b^10)*d*cos(d*x + c))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*sec(d*x+c)**2/(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.25187, size = 555, normalized size = 1.38 \begin{align*} \frac{\frac{{\left (20 \, a^{4} b^{3} - 7 \, a^{2} b^{5} + 2 \, b^{7}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{{\left (a^{9} - 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} - a^{3} b^{6}\right )} \sqrt{a^{2} - b^{2}}} + \frac{2 \,{\left (3 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{3} - 3 \, a b^{2}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}} + \frac{11 \, a^{3} b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, a b^{7} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 10 \, a^{4} b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 17 \, a^{2} b^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 6 \, b^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 29 \, a^{3} b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 8 \, a b^{7} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 10 \, a^{4} b^{4} - 3 \, a^{2} b^{6}}{{\left (a^{9} - 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} - a^{3} b^{6}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a\right )}^{2}} + \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)*sec(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

((20*a^4*b^3 - 7*a^2*b^5 + 2*b^7)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) +
b)/sqrt(a^2 - b^2)))/((a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*sqrt(a^2 - b^2)) + 2*(3*a^2*b*tan(1/2*d*x + 1/2*
c) + b^3*tan(1/2*d*x + 1/2*c) - a^3 - 3*a*b^2)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(tan(1/2*d*x + 1/2*c)^2 -
1)) + (11*a^3*b^5*tan(1/2*d*x + 1/2*c)^3 - 4*a*b^7*tan(1/2*d*x + 1/2*c)^3 + 10*a^4*b^4*tan(1/2*d*x + 1/2*c)^2
+ 17*a^2*b^6*tan(1/2*d*x + 1/2*c)^2 - 6*b^8*tan(1/2*d*x + 1/2*c)^2 + 29*a^3*b^5*tan(1/2*d*x + 1/2*c) - 8*a*b^7
*tan(1/2*d*x + 1/2*c) + 10*a^4*b^4 - 3*a^2*b^6)/((a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*(a*tan(1/2*d*x + 1/2*
c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2) + log(abs(tan(1/2*d*x + 1/2*c)))/a^3)/d